Answer
$$\frac{1}{{\sqrt 2 }}ta{n^{ - 1}}\left( {\frac{{x - 1}}{{\sqrt 2 }}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{2 + {{\left( {x - 1} \right)}^2}}}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = x - 1,{\text{ so that }}du = dx \cr
& {\text{then}} \cr
& \int {\frac{{dx}}{{2 + {{\left( {x - 1} \right)}^2}}}} = \int {\frac{{du}}{{2 + {u^2}}}} \cr
& = \int {\frac{{du}}{{\left( {\sqrt 2 } \right) + {u^2}}}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = \sqrt 2 \cr
& = \frac{1}{{\sqrt 2 }}ta{n^{ - 1}}\left( {\frac{u}{{\sqrt 2 }}} \right) + C \cr
& {\text{write in terms of }}x;{\text{ replace }}x - 1{\text{ for }}u \cr
& = \frac{1}{{\sqrt 2 }}ta{n^{ - 1}}\left( {\frac{{x - 1}}{{\sqrt 2 }}} \right) + C \cr} $$
